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## Differential Calculus

### Course: Differential Calculus>Unit 1

Lesson 16: Intermediate value theorem

# Justification with the intermediate value theorem: equation

Example justifying use of intermediate value theorem (where function is defined with an equation).

## Video transcript

- [Instructor] Let g of x equal one over x. Can we use the intermediate value theorem to say that there is a value c, such that g of c is equal to zero, and negative one is less than or equal to c, is less than or equal to one? If so, write a justification. So in order to even use the intermediate value theorem, you have to be continuous over the interval that you care about. And this interval that we care about is from x equals negative one to one. And, one over x is not continuous over that interval, it is not defined when x is equal to zero. And so, we could say, no, because, g of x not defined, or I could say not continuous. It's also not defined on every point of the interval, but let's say not continuous over the closed interval from negative one to one. And we could even put in parentheses not defined, at x is equal to zero. All right, now let's start asking the second question. Can we use the intermediate value theorem to say that the equation g of x is equal to 3/4 has a solution where 1 is less than or equal to x, is less than or equal to two? If so, write a justification. All right so first let's look at the interval. If we're thinking about the interval from one to two, well, yeah, our function is going to be continuous over that interval, so we could say g of x is continuous on the closed interval from one to two. And if you wanted to put more justification there, you could say g defined for all real numbers, such that x does not equal zero. I could write g of x defined for all real numbers such that x does not equal to zero, and you could say rational functions like one over x, are continuous at all points in their domains. That's going really establishing that g of x is continuous on that interval. And then we wanna see what values does g take over, take on at the end point, or actually, these are the end points we're looking at right over here. G of one is going to be equal to one over one is one, and g of two is going to be one over two. So, 3/4 is between g of one and g of two, so by the intermediate value theorem, there must be an x that is in the interval from where it's talking about the interval from one to two, such that g of x is equal to 3/4. And so, yes, we can use the intermediate value theorem to say that the equation g of x is equal to 3/4 has a solution, and we are done.